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Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments, Wiley, NY

Chapter 9 Three-Level and Mixed-Level Factorial and Fractional Factorial Design

Solutions 9.1. The effects of developer strength (A) and developer time (B) on the density of photographic plate film are being studied. Three strengths and three times are used, and four replicates of a 32 factorial experiment are run. The data from this experiment follow. Analyze the data using the standard methods for factorial experiments.

Developer Strength 1 2 3

10 0 5 4 7 7 8

Design Expert Output Response: Data ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean Source Squares DF Square Model 224.22 8 28.03 A 198.22 2 99.11 B 22.72 2 11.36 AB 3.28 4 0.82 Pure Error 71.00 27 2.63 Cor Total 295.22 35

Development Time (minutes) 14 2 1 3 2 4 4 2 4 6 6 8 9 5 7 7 8 10 10 10 12 7 8 7 9

F Value 10.66 37.69 4.32 0.31

Prob > F < 0.0001 < 0.0001 0.0236 0.8677

18 5 6 10 5 10 8

significant

The Model F-value of 10.66 implies the model is significant. There is only a 0.01% chance that a "Model F-Value" this large could occur due to noise. Values of "Prob > F" less than 0.0500 indicate model terms are significant. In this case A, B are significant model terms.

Strength and time are significant. The quadratic and interaction effects are not significant. By treating both A and B as numerical factors, the analysis can be performed as follows: Design Expert Output Response: Data ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean Source Squares DF Square Model 214.71 2 107.35 A 192.67 1 192.67 B 22.04 1 22.04 Residual 80.51 33 2.44 Lack of Fit 9.51 6 1.59 Pure Error 71.00 27 2.63 Cor Total 295.22 35

F Value 44.00 78.97 9.03

Prob > F < 0.0001 < 0.0001 0.0050

0.60

0.7255

The Model F-value of 44.00 implies the model is significant. There is only a 0 01% chance that a "Model F-Value" this large could occur due to noise

significant

not significant

Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments, Wiley, NY 9.2.

Compute the I and J components of the two-factor interaction in Problem 9.1. B 10 28 35

11 22 32

A

AB Totals = 77, 78, 71; I ( AB) = 2 AB Totals = 78, 74, 74; J ( AB ) =

17 32 39

77 2 + 78 2 + 71 2 22 6 2 − = 2.39 12 36 78 2 + 74 2 +74 2 22 6 2 − = 0.89 12 36

SS AB = I (AB ) + J ( AB ) = 3.28 9.3. An experiment was performed to study the effect of three different types of 32-ounce bottles (A) and three different shelf types (B) -- smooth permanent shelves, end-aisle displays with grilled shelves, and beverage coolers -- on the time it takes to stock ten 12-bottle cases on the shelves. Three workers (factor C) were employed in this experiment, and two replicates of a 33 factorial design were run. The observed time data are shown in the following table. Analyze the data and draw conclusions.

Worker 1

2

3

Bottle Type Plastic 28-mm glass 38-mm glass Plastic 28-mm glass 38-mm glass Plastic 28-mm glass 38-mm glass

Permanent 3.45 4.07 4.20 4.80 4.52 4.96 4.08 4.30 4.17

Replicate I End Aisle 4.14 4.38 4.26 5.22 5.15 5.17 3.94 4.53 4.86

Design Expert Output Response: Time ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean Source Squares DF Square Model 28.28 26 1.09 A 0.41 2 0.21 B 17.75 2 8.88 C 7.66 2 3.83 AB 0.12 4 0.029 AC 0.11 4 0.027 BC 1.68 4 0.42 ABC 0.55 8 0.069 Pure Error 2.03 27 0.075 Cor Total 30.31 53

Cooler 5.80 5.48 5.67 6.21 6.25 6.03 5.14 4.99 4.85

F Value 14.50 2.74 118.34 51.09 0.39 0.36 5.60 0.92

Prob > F < 0.0001 0.0828 < 0.0001 < 0.0001 0.8163 0.8319 0.0021 0.5145

The Model F-value of 14.50 implies the model is significant. There is only a 0.01% chance that a "Model F-Value" this large could occur due to noise. V l

f "P b

F" l

th

0 0500 i di t

d lt

i

Permanent 3.36 3.52 3.68 4.40 4.44 4.39 3.65 4.04 3.88

ifi

t

Replicate II End Aisle 4.19 4.26 4.37 4.70 4.65 4.75 4.08 4.08 4.48

significant

Cooler 5.23 4.85 5.58 5.88 6.20 6.38 4.49 4.59 4.90

Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments, Wiley, NY

Factors B and C, shelf type and worker, and the BC interaction are significant. For the shortest time regardless of worker chose the permanent shelves. This can easily be seen in the interaction plot below.

9.4. A medical researcher is studying the effect of lidocaine on the enzyme level in the heart muscle of beagle dogs. Three different commercial brands of lidocaine (A), three dosage levels (B), and three dogs (C) are used in the experiment, and two replicates of a 33 factorial design are run. The observed enzyme levels follow. Analyze the data from this experiment.

Lidocaine Brand 1

2

3

Dosage Strength 1 2 3 1 2 3 1 2 3

1 96 94 101 85 95 108 84 95 105

Replicate I Dog 2 84 99 106 84 98 114 83 97 100

Design Expert Output Response: Enzyme Level ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean Source Squares DF Square Model 4346.26 26 167.16 A 31.37 2 15.69 B 4040.04 2 2020.02 C 25.04 2 12.52 AB 112.52 4 28.13

3 85 98 98 86 97 109 81 93 106

F Value 13.10 1.23 158.32 0.98 2.20

1 84 95 105 80 93 110 83 92 102

Prob > F < 0.0001 0.3083 < 0.0001 0.3879 0.0952

Replicate II Dog 2 85 97 104 82 99 102 80 96 111

significant

3 86 90 103 84 95 100 79 93 108

Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments, Wiley, NY Pure Error Cor Total

344.50 4690.76

27 53

12.76

The Model F-value of 13.10 implies the model is significant. There is only a 0.01% chance that a "Model F-Value" this large could occur due to noise. Values of "Prob > F" less than 0.0500 indicate model terms are significant. In this case B are significant model terms.

The factor B, dosage strength, is significant.

9.5.

Compute the I and J components of the two-factor interactions for Problem 9.4.

B

520 573 617

A 501 577 643

490 566 632

I totals = 1684, 1729, 1706 J totals = 1729, 1684, 1706 I(AB) = 56.259 J(AB) = 56.259 SSAB = 112.519

C

575 575 560

A 571 579 571

561 567 560

I totals = 1700, 1713, 1706 J totals = 1714, 1698, 1707 I(AC) = 4.704 J(AC) = 7.148 SSAC = 11.852

C

512 498 501

B 564 586 566

631 637 624

I totals = 1718, 1715, 1686 J totals = 1722, 1702, 1695 I(BC) = 34.704 J(BC) = 21.815 SSBC = 56.519 9.6. An experiment is run in a chemical process using a 32 factorial design. The design factors are temperature and pressure, and the response variable is yield. The data that result from this experiment are shown below.

Temperature, °C 80 90 100

(a)

100 47.58, 48.77 51.86, 82.43 71.18, 92.77

Pressure, psig 120 64.97, 69.22 88.47, 84.23 96.57, 88.72

140 80.92, 72.60 93.95, 88.54 76.58, 83.04

Analyze the data from this experiment by conducting an analysis of variance. What conclusions can you draw?

Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments, Wiley, NY Design Expert Output Response: Yield ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Source Squares DF Model 3187.13 8 A 1096.93 2 B 1503.56 2 AB 586.64 4 Pure Error 819.98 9 Cor Total 4007.10 17

Mean Square 398.39 548.47 751.78 146.66 91.11

F Value 4.37 6.02 8.25 1.61

Prob > F 0.0205 0.0219 0.0092 0.2536

significant

The Model F-value of 4.37 implies the model is significant. There is only a 2.05% chance that a "Model F-Value" this large could occur due to noise. Values of "Prob > F" less than 0.0500 indicate model terms are significant. In this case A, B are significant model terms.

Temperature and pressure are significant. Their interaction is not. An alternate analysis is performed below with A and B treated as numeric factors: Design Expert Output Response: Yield ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Source Squares DF Model 3073.27 5 614.65 A 850.76 1 850.76 B 1297.92 1 1297.92 A2 246.18 1 246.18 B2 205.64 1 205.64 AB 472.78 1 472.78 Residual 933.83 12 77.82 Lack of Fit 113.86 3 37.95 Pure Error 819.98 9 91.11 Cor Total 4007.10 17

Mean Square 7.90 10.93 16.68 3.16 2.64 6.08 0.42

F Value 0.0017 0.0063 0.0015 0.1006 0.1300 0.0298

Prob > F significant

0.7454 not significant

The Model F-value of 7.90 implies the model is significant. There is only a 0.17% chance that a "Model F-Value" this large could occur due to noise. Values of "Prob > F" less than 0.0500 indicate model terms are significant. In this case A, B, AB are significant model terms.

(b)

Graphically analyze the residuals. Are there any concerns about underlying assumptions or model adequacy?

The following residual plots are based on the first analysis shown above with the A and B treated as categorical factors. The plot of residuals versus pressure shows a decreasing funnel shape indicating a non-constant variance.

Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments, Wiley, NY

(c)

Verify that if we let the low, medium and high levels of both factors in this experiment take on the levels -1, 0, and +1, then a least squares fit to a second order model for yield is

ˆy = 86.81 + 10.4 x1 + 8.42 x2 − 7.17 x21 − 7.84 x22 − 7.69x1 x2 The coefficients can be found in the following table of computer output. Design Expert Output Final Equation in Terms of Coded Factors: Yield = +86.81 +8.42 +10.40 -7.84

*A *B * A2

-7.17 -7.69

* B2 *A*B

Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments, Wiley, NY (d)

Confirm that the model in part (c) can be written in terms of the natural variables temperature (T) and pressure (P) as

=−

+

+

−

−

−

The coefficients can be found in the following table of computer output. Design Expert Output Final Equation in Terms of Actual Factors: Yield -1335.62500 +8.58737 +18.55850

(e)

=

-0.019612

* Pressure * Temperature * Pressure2

-0.071700 -0.038437

* Temperature2 * Pressure * Temperature

Construct a contour plot for yield as a function of pressure and temperature. Based on the examination of this plot, where would you recommend running the process?

Run the process in the oval region indicated by the yield of 90.

Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments, Wiley, NY 9.7. (a)

Confound a 33 design in three blocks using the ABC2 component of the three-factor interaction. Compare your results with the design in Figure 9.7. L = x1 + x 2 + 2x 3 Block 1 000 112 210 120 022 202 221 101 011

Block 2 100 212 010 220 122 002 021 201 111

Block 3 200 012 110 020 222 102 121 001 211

The new design is a 180° rotation around the Factor B axis. (b)

Confound a 33 design in three blocks using the AB2C component of the three-factor interaction. Compare your results with the design in Figure 9.7. L = x1 + 2x2 + x3 Block 1 000 022 011 212 220 201 110 102 121

Block 2 210 202 221 100 122 111 012 020 001

Block 3 112 120 101 010 002 021 200 222 211

The new design is a 180° rotation around the Factor C axis. (c)

Confound a 33 design three blocks using the ABC component of the three-factor interaction. Compare your results with the design in Figure 9.7. L = x1 + x2 + x3 Block 1 000 210 120 021 201 111 012 222 102

h

d i

i

90

i

d h

Block 2 112 022 202 100 010 220 121 001 211

Block 3 221 101 011 212 122 002 200 110 020

i

l

ih

i hi

l

0

dl

1i

Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments, Wiley, NY (d)

After looking at the designs in parts (a), (b), and (c) and Figure 9.7, what conclusions can you draw?

All four designs are relatively the same. The only differences are rotations and swapping of layers. 9.8. Confound a 34 design in three blocks using the AB2CD component of the four-factor interaction. The three blocks are shown below with L = x1 + 2x2 + x3 + x4

0000 0211 2102

1100 1222 0021

0110 2212 2001

0101 2221 2120

Block 1 2200 0122 1011

0220 2111 2022

0202 1121 0012

1210 1112 1002

1201 2010 1020

1021 0200 1000

1110 0022 1122

1202 0111 1211

0001 2002 2112

Block 2 0120 2121 2201

0212 2210 2020

1012 0010 2011

1101 0102 2100

1220 0221 2222

2012 1221 2021

2101 1010 2110

2220 1102 2202

1022 0020 0100

Block 3 1111 0112 0222

1200 0201 0011

2000 1001 0002

2121 1120 0121

2211 1212 0210

9.9. Consider the data from the first replicate of Problem 9.3. Assuming that all 27 observations could not be run on the same day, set up a design for conducting the experiment over three days with AB2C confounded with blocks. Analyze the data.

000 110 011 102 201 212 121 022 220 Totals

Block 1 = 3.45 = 4.38 = 5.22 = 4.30 = 4.96 = 4.86 = 6.25 = 5.14 = 5.67 = 44.23

100 210 111 202 001 012 221 122 020

Block 2 = = = = = = = = =

4.07 4.26 5.15 4.17 4.80 3.94 6.03 4.99 5.80 43.21

200 010 211 002 101 112 021 222 120

Block 3 = 4.20 = 4.14 = 5.17 = 4.08 = 4.52 = 4.53 = 6.21 = 4.85 = 5.48 43.18

The analysis of variance below identifies factors B and C as significant. Design Expert Output Response: Time ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Source Squares DF Block 0.079 2 Model 13.57 18 A 0.11 2 B 8.42 2 C 3 81 2

Mean Square 0.040 0.75 0.055 4.21 1 91

F Value 10.49 0.77 58.63 26 53

Prob > F 0.0040 0.5052 0.0001 0 0010

significant

Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments, Wiley, NY BC Residual Cor Total

0.81 0.43 14.08

4 6 26

0.20 0.072

2.83

0.1234

The Model F-value of 10.49 implies the model is significant. There is only a 0.40% chance that a "Model F-Value" this large could occur due to noise. Values of "Prob > F" less than 0.0500 indicate model terms are significant. In this case B, C are significant model terms.

9.10. Outline the analysis of variance table for the 34 design in nine blocks. Is this a practical design? Source A B C D AB AC AD BC BD CD ABC (AB2C,ABC2,AB2C2) ABD (ABD,AB2D,ABD2) ACD (ACD,ACD2,AC2D2) BCD (BCD,BC2D,BCD2) ABCD Blocks (ABC,AB2C2,AC2D,BC2D2) Total

DF 2 2 2 2 4 4 4 4 4 4 6 6 6 6 16 8 80

Any experiment with 81 runs is large. Instead of having three full levels of each factor, if two levels of each factor could be used, then the overall design would have 16 runs plus some center points. This two-level design could now probably be run in 2 or 4 blocks, with center points in each block. Additional curvature effects could be determined by augmenting the experiment with the axial points of a central composite design and additional enter points. The overall design would be less than 81 runs.

9.11. Consider the data in Problem 9.3. If ABC is confounded in replicate I and ABC2 is confounded in replicate II, perform the analysis of variance. L1 = x 1 + x 2 + x 3 Block 1 000 = 3.45 111 = 5.15 222 = 4.85 120 = 5.48 102 = 4.30 210 = 4.26 201 = 4.96 012 = 3.94 021 = 6.21

Block 2 001 = 4.80 112 = 4.53 220 = 5.67 121 = 6.25 100 = 4.07 211 = 5.17 202 = 4.17 010 = 4.14 022 = 5.14

L2 = x1 + x2 + 2x2 002 110 221 122 101 212 200 011 020

Block 3 = 4.08 = 4.38 = 6.03 = 4.99 = 4.52 = 4.86 = 4.20 = 5.22 = 5.80

000 101 011 221 202 022 120 210 112

Block 1 = 3.36 = 4.44 = 4.70 = 6.38 = 3.88 = 4.49 = 4.85 = 4.37 = 4.08

100 201 111 021 002 122 220 010 212

Block 2 = 3.52 = 4.39 = 4.65 = 5.88 = 3.65 = 4.59 = 5.58 = 4.19 = 4.48

200 001 211 121 102 222 020 110 012

Block 3 = = = = = = = = =

3.68 4.40 4.75 6.20 4.04 4.90 5.23 4.26 4.08

The sums of squares for A, B, C, AB, AC, and BC are calculated as usual. The only sums of squares presenting difficulties with calculations are the four components of the ABC interaction (ABC ABC2

Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments, Wiley, NY We will show how to calculate AB2C and AB2C2 from both replicates. Form a two-way table of A x B at each level of C. Find the I(AB) and J(AB) totals for each third of the A x B table.

C 0

1

2

B 0 1 2 0 1 2 0 1 2

0 6.81 8.33 11.03 9.20 9.92 12.09 7.73 8.02 9.63

A 1 7.59 8.64 10.33 8.96 9.80 12.45 8.34 8.61 9.58

2 7.88 8.63 11.25 9.35 9.92 12.41 8.05 9.34 9.75

I 26.70 27.25 26.54 31.41 30.97 31.72 26.09 27.31 25.65

J 27.55 27.17 25.77 31.24 31.29 31.57 26.29 26.11 26.65

The I and J components for each third of the above table are used to form a new table of diagonal totals. C 0 1 2

26.70 31.41 26.09

I(AB) 27.25 30.97 27.31

26.54 31.72 25.65

J(AB) 27.17 31.29 26.11

27.55 31.24 26.29

25.77 31.57 26.65

I Totals: 85.06, 85.26, 83.32

I Totals: 85.49, 85.03, 83.12

J Totals: 85.73, 83.60, 84.31

J Totals: 83.35,8 5.06, 85.23

(85.06) 2 + (85.26) 2 +(83.32) 2 (253.64) 2 − =0.1265 18 54 (85.73) 2 + (83.60) 2 + (84.31 )2 (253.64)2 − = 0.1307 and, AB 2C = J ⎡⎣CxI (AB )⎤⎦ = 18 54 Now, AB2C 2 = I ⎡⎣CxI (AB )⎤⎦ =

If it were necessary, we could find ABC2 as ABC2= I[C x J(AB)] and ABC as J[C x J(AB)]. However, these components must be computed using the data from the appropriate replicate. The analysis of variance table: Source Replicates Blocks within Replicates A B C AB AC BC ABC (rep I) ABC2 (rep II) AB2C AB2C2 Error Total

SS 1.06696 0.2038 0.4104 17.7514 7.6631 0.1161 0.1093 1.6790 0.0452 0.1020 0.1307 0.1265 0.8998 30.3069

DF 1 4 2 2 2 4 4 4 2 2 2 2 22 53

MS

0.2052 8.8757 3.8316 0.0290 0.0273 0.4198 0.0226 0.0510 0.0754 0.0633 0.0409

F0

5.02 217.0 93.68 F < 0.0001 0.1862 0.0002 < 0.0001 0.9331 0.0334 0.9617 0.6996

significant

The Model F-value of 4.89 impli...